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Sliva [168]
3 years ago
8

A sailor on a trans-Pacific solo voyage notices one day that if he puts 375. mL of fresh water into a plastic cup fresh water we

ighing 15.0 g, the cup floats in the seawater around his boat with the fresh water inside the cup at exactly the same level as the seawater outside the cup.
Required:
Calculate the amount of salt dissolved in each liter of seawater.
Chemistry
1 answer:
ElenaW [278]3 years ago
3 0

Answer:

Amount of salt dissolved in each liter of seawater = 40 g

Explanation:

According to Archimedes principle, a body will float in a fluid if the upthrust experienced by a body is equal to the to the weight of the body.

Also, the volume of seawater displaced equals the volume of freshwater in the cup.

From the above principle, since the freshwater and cup floats in the seawater, their combined weight equals the upthrust.

Therefore, mass of equal volume of displaced seawater = mass of freshwater + mass of cup

Mass of freshwater = density of freshwater * volume

density of freshwater = 1 g/mL; volume = 375 mL

mass of freshwater = 375 mL * 1 g/mL = 375 g

mass of seawater = 375 + 15 = 390 g

mass of salt in 375 mL seawater = mass of seawater - mass of freshwater

mass of salt = (390 - 375) g = 15 g

Since 15 g of salt are dissolved in 375 mL seawater, mass of salt in 1 L of seawater =(1000 mL/ 375) * 15g = 40 g

Therefore, amount of salt dissolved in each liter of seawater = 40 g

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Describe the properties of ammonium lauryl sulfate that make it a feasible surfactant
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If 18.1 g of ammonia is added to 27.2 g of oxygen gas, how many grams of excess reactant is remaining once the reaction has gone
GREYUIT [131]

Answer:

m of NH3 = 6.46 g

Explanation:

First, in order to know the limiting and excess reactant, we need to write and balance the equation that is taking place:

NH₃ + O₂ ---------> NO + H₂O

Now, let's balance the equation:

4NH₃ + 5O₂ ---------> 4NO + 6H₂O

Now that we have the balanced equation, let's see which reactant is in excess. To know that, let's calculate the moles of each reactant using the molar mass:

MM NH3 = 17 g/mol

MM O2 = 32 g/mol

moles NH3 = 18.1 / 17 = 1.06 moles

moles O2 = 27.2 / 32 = 0.85 moles

Now, let's compare these moles with the theorical moles that the balanced equation gave:

4 moles NH3 --------> 5 moles O2

1.06 moles ----------> X

X = 1.06 * 5 / 4 = 1.325 moles of O2

These means in order to  NH3 completely reacts with O2, it needs 1.325 moles of O2, which we don't have it. We only have 0.85 moles of O2, therefore, the limiting reactant is the O2 and the excess is NH3.

Now, let's see how many grams in excess we have left after the reaction is complete.

4 moles NH3 --------> 5 moles O2

X moles NH3 ----------> 0.85 moles

X = 0.85 * 4 / 5 = 0.68 moles of NH3

This means that 0.85 moles of O2 will react with only 0.68 moles of NH3, and we have 1.06 so, the remaining moles are:

moles remaining of NH3 = 1.06 - 0.68 = 0.38 moles

Finally the mass:

m = 0.38 * 17

<em>m = 6.46 g of NH3</em>

8 0
3 years ago
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